My multiple-variable-calculus is not so strong so the following problem got me stuck.

I have density function

f(x,y) = x^2 + xy/3 for 0<x<1; 0<y<2 otherwise 0

And I need to calculate Prob(X > Y). X and Y are random variables.

I know how to do Prob(X <= 0.5) etc.

Just to make sure, this is just integrating over all values of Y, such that X <= 0.5, which is a rectangle.

Niels said:

Also would be nice if someone could explain Prob(Y < 1/2 and X < 1/2)

Similarly, this is an integration of the density function over the region where Y < 1/2 and X < 1/2. This is a square of sidelength 1/2 with one corner at the origin. It is a double integral that can be written as:
[tex]\int_0^{\frac{1}{2}} \int_0^{\frac{1}{2}} f(x,y) dy dx[/tex]
Do you know how to find P(X<1/2 OR Y<1/2) ?

Niels said:

and
Prob(X+Y < 1)

If you cannot picture the region, rewrite it in a friendlier form; ie., Y < -X + 1. You're then integrating the density function for all points below the line y = -x + 1. This integral can be written:
[tex]\int_0^1 \int_0^{-x+1} f(x,y) dy dx[/tex]
So your original problem, P(Y<X) is just the set of points below the line y=x.